Using the probability table, it is found that:
- a) There is a 0.25 = 25% probability that this couple spends 45 dollars or more.
- b) The expected amount the couple actually has to pay is $36.85.
Item a:
To find the probabilities involving the total cost, we have to <u>add the variables X and Y</u> from the table, then:





The probability involving <u>values of 45 or more</u> is:

0.25 = 25% probability that this couple spends 45 dollars or more.
Item b:
For a <em>discrete distribution</em>, the expected value is the <u>sum of each outcome multiplied by it's respective probability</u>, hence, involving the 10% discount for prices above $45:
![E(X) = 0.2(30) + 0.3(35) + 0.25(40) + 0.9[0.2(45) + 0.05(50)] = 36.85](https://tex.z-dn.net/?f=E%28X%29%20%3D%200.2%2830%29%20%2B%200.3%2835%29%20%2B%200.25%2840%29%20%2B%200.9%5B0.2%2845%29%20%2B%200.05%2850%29%5D%20%3D%2036.85)
The expected amount the couple actually has to pay is $36.85
A similar problem is given at brainly.com/question/25782059
Answer:
4 is included in the domain
Step-by-step explanation:
Answer:
a) 18
b) 2.101
Step-by-step explanation:
a) What are the degrees of freedom for Student's t distribution when the sample size is 19?
Degrees of freedom = n - 1
Where n = sample size
= 19 - 1
= 18
b) Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 19.
(Round your answer to three decimal places.)
We would be determining these using the t distribution table
1 - 0.95 = 0.05 ( for two tailed)
Or
0.05/2 = 0.025(one - tailed)
Hence, the tc(test score) for a 0.95 confidence level when the sample is 19 is 2.101
A.) rectangle would be the correct answer